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Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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AH°F (Na" (aq)) = -240.1 kJ/mol AH°rxn =2 AH°f (products) - E AH°r (reactants) %3D AH°rxn = [(1 mol x AH°r (CH3COOH(1)) +( 1 mol x AH°; (Na2CO3(s))] - [(1 mol x AH°r (CH3CO,Na(aq)) + (1 mol x AHr (CO2(g)) + (1 mol x AH°f (H2O(1))+ (1 mol x AH°f (Na* (aq))] AH°rxn = [ (-484.5 kJ/mol) + (-1130.7 kJ/mol)] - [1 mol x (-726.1 kJ/mol) + (-393.5 kJ/mol) + (-241.8 kJ/mol) + (-240.1 kJ/mol] %3D AH rxn = - 13.7 kJ Therefore 10.00 L of acetic acid produces -13.7 kJ of heat when reacted with Na,CO3(s) MacBook Air
Transcribed Image Text:AH°F (Na" (aq)) = -240.1 kJ/mol AH°rxn =2 AH°f (products) - E AH°r (reactants) %3D AH°rxn = [(1 mol x AH°r (CH3COOH(1)) +( 1 mol x AH°; (Na2CO3(s))] - [(1 mol x AH°r (CH3CO,Na(aq)) + (1 mol x AHr (CO2(g)) + (1 mol x AH°f (H2O(1))+ (1 mol x AH°f (Na* (aq))] AH°rxn = [ (-484.5 kJ/mol) + (-1130.7 kJ/mol)] - [1 mol x (-726.1 kJ/mol) + (-393.5 kJ/mol) + (-241.8 kJ/mol) + (-240.1 kJ/mol] %3D AH rxn = - 13.7 kJ Therefore 10.00 L of acetic acid produces -13.7 kJ of heat when reacted with Na,CO3(s) MacBook Air
Page < 2 > of 3 ZOOM + Question 1 Acid spills are often neutralized with solid sodium carbonate. The products of these reactions include CO2(g) and H20(1). a) How many kilograms of solid sodium carbonate is needed to neutralize a 10.00L spill of concentrated acetic acid? (density = 1.049g/mL) %3D b) How much heat ( in kilojoules) is absorbed or liberated in the neutralization of 10.00L of acetic acid with Na,CO,? (the AH°; (CH;CO,Na(aq)) = -726.1 kJ/mol) Balance equation: CH;COOH(1) + Na,CO;(s) → CH;COONA(aq) + CO2(g) + H,O(1) + Na*(aq) a) 10.00 L CH3COOH(1) x 1.049 g / mL = 10.490 g CH3COOH 1 mol CH3COOH Moles of acetic acid = 10.490 g CH3COOH (1) x 0.17468 mol of CH;COOH 60.052 g CH3COOH 1 mol Na2CO3 Moles of Na,C03(s) = 0.17468 mol CH3COOH x 1 mol CH3C00H = 0.17468 mol of Na,CO, Mass of Na,CO3(s) = 0.1746g mol of Na,CO3(s) x- 82.998 g Na2CO3 1 mol NA2CO3 14.5 kg Na2CO3 needed to neutralize 10.00 L of acetic acid. b) Need to find AHPxn... AH° (CH3COOH(1)) = -484.5 kJ/mol AH°F (CH3CO2Na(aq)) = -726.1 kJ/mol AH°F (Na2CO3(s)) = -1130.7 kJ/mol AH°; (CO2(g)) = -393.5 kJ/mol AH°F (H2O(1)) = -241.8 kJ/mol AH°F (Na* (aq)) = -240.1 kJ/mol MacBock Air wwwwcn
Transcribed Image Text:Page < 2 > of 3 ZOOM + Question 1 Acid spills are often neutralized with solid sodium carbonate. The products of these reactions include CO2(g) and H20(1). a) How many kilograms of solid sodium carbonate is needed to neutralize a 10.00L spill of concentrated acetic acid? (density = 1.049g/mL) %3D b) How much heat ( in kilojoules) is absorbed or liberated in the neutralization of 10.00L of acetic acid with Na,CO,? (the AH°; (CH;CO,Na(aq)) = -726.1 kJ/mol) Balance equation: CH;COOH(1) + Na,CO;(s) → CH;COONA(aq) + CO2(g) + H,O(1) + Na*(aq) a) 10.00 L CH3COOH(1) x 1.049 g / mL = 10.490 g CH3COOH 1 mol CH3COOH Moles of acetic acid = 10.490 g CH3COOH (1) x 0.17468 mol of CH;COOH 60.052 g CH3COOH 1 mol Na2CO3 Moles of Na,C03(s) = 0.17468 mol CH3COOH x 1 mol CH3C00H = 0.17468 mol of Na,CO, Mass of Na,CO3(s) = 0.1746g mol of Na,CO3(s) x- 82.998 g Na2CO3 1 mol NA2CO3 14.5 kg Na2CO3 needed to neutralize 10.00 L of acetic acid. b) Need to find AHPxn... AH° (CH3COOH(1)) = -484.5 kJ/mol AH°F (CH3CO2Na(aq)) = -726.1 kJ/mol AH°F (Na2CO3(s)) = -1130.7 kJ/mol AH°; (CO2(g)) = -393.5 kJ/mol AH°F (H2O(1)) = -241.8 kJ/mol AH°F (Na* (aq)) = -240.1 kJ/mol MacBock Air wwwwcn
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